Pictured at ~490C on the thermocouple. True color was not as bright and strongly orange.

After the earlier success of producing apparent excess heating with a blend of nickel powder, lithium hydroxide, aluminum powder, and iron oxide (Fe2O3), there followed three failed attempts to repeat the results. The present experiment appears to have been successful. The fuel was modified in this experiment thanks to fellow Vortex member Jones Beene who sent a special ceramic powder mix based on lime. It was believed that the special luminescence properties of lime may enhance the effect. The fuel was composed primarily of the lime-based powder by volume. Another difference with the present experiment is that the entire tube was filled with the fuel. The actual nickel weight is significantly less than used in the Lugano test, but I don’t have precise measurements on the amount.

The current experiment utilized a second set of temperature sensors (one for measuring ambient temperature, and the other attached to the top of the tube furnace). This allows the tube furnace to act as a simple calorimeter and removes the need for a control run to be performed with every experiment.

The experiment generated a maximum of 10.6 watts of excess heating over 90 watts input (11.8% gain). Of note was that the excess heating was much more persistent.

Chart A. The experimental run stepped from 10, 50, 90W of input. The calibration run stepped from 10, 30, 50, 70, and 90W of input.

Note that the excess heating does not occur until the 90W step.

Chart B. Showing calculated excess power from the time it started over a period of [insert]

Because the excess heating was of long duration, and showed no signs of stopping, some steps were taken to attempt to quench the reaction (assuming that is the cause). The tube furnace was opened and the cell photographed. The resulting temperature drop was quickly reversed. The next attempt to quench the reaction was performed by stepping the power down by 20W per step. Each step was allowed to run for approximately 4 hours for the temperature to settle. There appeared to be excess heat at every step until 10W was reached. At the 10W level there was no indication of excess heating. When the power input was again increased to 50W, the apparent excess heating returned.

Chart C. Temperatures of the external wall of the reactor tube. Note that the internal temperatures are actually much higher, but not measured. The power steps from 10, 50, 90, 70, 50, 30, 10, 50, 90 watts.

Conclusions:

Two out of five experiments demonstrated apparent excess heating with a maximum of ~10W. Assuming there is no systematic factor that has been missed in these experiments, it may indicate LiOH and aluminum powder as viable alternatives to the more dangerous LAH used in the Rossi and Parkhomov formulations.

The experiment continues to run, and updates will be reported here.

Update:

After attempting a higher power input (130W) the resistance wire broke ending the experiment. Here is a chart of the calculated excess running over the 90W step lasting 19 hours and 24 minutes.

Additionally, here is a chart showing calculated excess power at each step in the experiment.

The results of this experiment are not definitive, but do show enough promise to warrant further investigation. The next steps will be working towards increasing the COP and improving calorimetry.

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I’m not sure that aging elements will actually produce more heat. They may glow more brightly because the diameter has been reduced and thus the resistance increased. The power dissipated in the circuit will remain the same with the adjustments from the power supply control software. I could be wrong about this, but it seems to be a matter of conservation of energy.

And Roy is correct that a big part of the battle is avoiding self-deception. I’ve made plenty of mistakes and have disproved what I thought may be excess heat in the past several times. We’ll see if this turns out to be real as the experiments continue. I will be convinced when a COP of 1.5 is achieved.

If your going the whole way it’s just as important to know what doesn’t work as what does work. The fact that everything we learn i.e. “aging of filament’s and heating elements” does effect the total output of the experiment and the way you have structured the arrangement means that these changes are isolatable.

Consistency is the most important factor in low cost experiments and even though your budget may not be equal to CERNS you can still do good science and you are on a good track as exampled in your work.

Your trying to avoid self deception and science,is merely a collection of techniques for doing that.

Every time I look I see something and all observable phenomena have not been harnessed yet, many observable phenomena remain in plain sight yet remain unseen.

I am encouraged with your work.
Roy sais “the #1 rule in life is to be sure and make enough mistakes, because if you aren’t making mistakes your not doing anything”.

Yes, very good recommendations there. The nickel powder attenuates the likelihood of explosion, but you better believe it can happen. Not only can you bit hit by pieces of the apparatus, but you will blow the contents of the reactor all over the room and into the air if it is not in some kind of enclosure.

I was meaning the ID of the outer metal tube. The two thermocouples used in the delta T calculation should be on opposite sides of a thin solid. The idea is that no matter how the heat is generated or transported internally it all must leave by conduction through this instrumented wall. You may find that the metal is too good a thermal conductor and therefore at your power levels you don’t need a large thermal gradient to dissipate the heat flux. In that case you could “wrap” your metal tube in a poor thermal conductor and use you existing outside thermocouple for one side and put the second on the outside of the wrap. Ie. Once again sandwiching the conductor layer.

Depending on your budget there are commercial heat flux “patches” based on exactly this same heat conduction physics. They are designed to be affixed to hot metal surfaces like boilers. The idea being that the relatively small patch does not distort the overall heat flux and through careful calibration can yield good overall measurement.

I think I understand now–thank you. It is an elegant solution. I can wrap a thin sheet of ceramic insulation around the OD of the tube furnace and put one TC on the inside and one on the outside of the insulation. Does it matter if the inner one is touching the metal on top of the tube furnace?

Actually, might be able to do this easier. If you see the temperature sensor in the picture above, it is encased in an stainless sleeve. If another of the same type was attached to the top of the existing one, wouldn’t you get the same effect with the conduction through the steel? I guess I would have to try it to see if there is enough of a thermal gradient.

No. In fact the metal of the tube furnace will conduct heat longittinally to even out things. The only thing you care about in the wrap solution is that the thermocouples are placed so that they measure delta T across a representative thickness compared to the rest of the tube. Ie. You want even radial heat conduction through the wrap. If you have the option to use thinner thermocouples under the wrap that would be better. The commercial sensors use these.

I have improved the setup based on your simulations for a conduction calorimeter. I have two temperature sensors with a layer of ceramic insulation sandwiched in between. Additionally, I put a layer of ceramic insulation over the bottom part of the tube furnace to help force the heat conduction upwards. The reaction chamber was also wrapped in ceramic insulation. Seems like the results are pretty good in a test run of the setup. The calibration line seems to be linear as your simulations predicted. Please take a look at this chart and see if you think it meets your expectations.

I’m not quite understanding how you wrapped the furnace tube. The key is that you want all the heat flux to exit radially through exactly the same thermal pathway as you instrumented with your thermocouple pair. The thermocouple pair don’t necessarily need to be exactly opposite each other but they do need to read the temp on each side of the wrap layer.

The ceramic insulation is wrapped over the top part of the tube furnace cylinder. Additionally, it is acting as a thermal barrier for the bottom part causing the heat convection and conduction upwards towards where the sensors are. See pictures.

In the first photo is that insulation around the reaction cylinder? For the conduction calorimeter to work you don’t need to worry about insulating the reaction chamber. The thermocouples look to be in a good position on the furnace tube.

It is worth clarifying that the purpose of the wrap is to provide a thin poorly conductive layer through which all the heat escapes. The reason for choosing a poor thermal conductor is to amplify the delta T for a given heat flux. The thermocouples are placed on a representative patch on this layer to measure the temperature drop as the heat is conducted across. The wrap should not be used to insulate and concentrate the heat flux into a particular area. It is best to allow the heat flux to be evenly distributed across all the furnace cylinder wall to give a more reliable conduction calorimeter. If you want to insulate anything do it at the caps top and bottom

Bob,
In the first photo, that is insulation around the reaction chamber. It is not to aid the calorimetry, but to reduce the amount of power input needed to reach higher temperatures.

I did understand your design from the simulation with the poor heat conductor being used allow the heat flux to be evenly distributed. It is truly an elegant solution and the easiest method of implementing calorimetry that I have seen.

It looks like it will work very well and give an error level of +/- 1W. It does indeed produce a linear equation as your simulations predicted.

Can you show a diagram of you thermocouple positions? Were they on oppsosite sides of a thin tube? The theory of a conduction calorimeter “requires” that the bulk of the heat flux go radially through a conduction pathway ie. Through the wall of the thermocouple instrumented tube. In that case a plot of delta T vs input power is predicted to be linear. Using this linear equation one can calculated excess power for any measured delta T

The type K thermocouple placement is shown in the picture. The temperature sensor that was used for the calculations was placed on top of the tube furnace allowing the furnace itself to act as a basic calorimeter (picture here). One possible limitation being that if the calibration run cell was in contact with the bottom of the inner ceramic cylinder that some heat might have been conducted in a different direction. I will improve this in the future with ceramic insulation to help limit the direction of the heat conduction.

It is attached to the outside of the alumina tube. The tube furnace is not actually used for heating (just housing the experiment). There are two more temp sensors back away from the tube furnace for measuring ambient temp.

Just add another thermocouple on the ID of the furnace tube opposite the OD one and the difference in temp btw the ID thermocouple and the OD thermocouple vs input power should yield the predicted linear relation during calibration. This linear relation can then be used to calculate any excess heat flux during actual run.

Sometimes LENR effects are delayed from onset of stimulus, so it is probably best to compare integrals of power over time. Ie. Cumulative KW-hr out vs in

Do you mean on the ID of the outer metal wall, or inside the tube part? There is an inner ceramic tube with heating elements (not used in this experiment). I’m not sure I can get to the inner side of the outer metal wall.

If you look at the picture at the top of the post, the upper part looks like the part shown in the picture.

Calcium carbonate will decompose into CO2 and CaO. At high enough temperatures, molecular H2 will decompose into atomic H. And even LiOH will decompose . These changes in gas composition will alter the thermal conductivity of the inner tube which will make the apparent tube temperature follow the conductivity changes. A stable gas mixture will give a stable temperature reading. I think better calorimetry would help sort out the simultaneous effects. There are some excellent high temp and pressure differential scanning calorimeters around. It would help remove the ambiguity.